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Concentration and Rate Relationship

Posted: Mon Mar 05, 2018 11:17 pm
by Amanda Mac 1C
In a second order reaction, if one reactant alone is doubled does that mean the rate is quadrupled? What happens if two reactants in a second order reaction are both doubled?

Re: Concentration and Rate Relationship

Posted: Mon Mar 05, 2018 11:41 pm
by Sean Monji 2B
I think so. This can be shown mathematically.
Lets say rate = k[A]^2
If A is doubled,
rate = k[2A]^2 = 4k[A]^2, meaning rate was quadrupled
in rate = k[A][B]
k[2A][2B] = 4k[A][B]
Though in the second reaction, if you only double A or B, the rate only doubles
k[2A][B] = 2k[A][B]
This has to do with how likely the molecules will collide to form the product.
Hope that makes sense

Re: Concentration and Rate Relationship

Posted: Tue Mar 06, 2018 12:18 am
by Adrian Lim 1G
For the most part, it would be easier to see the changes if you had the rate of reaction and the corresponding orders. From there, you could just plug in new concentrations to see what kind of effect it would have on the overall rate.

Re: Concentration and Rate Relationship

Posted: Tue Mar 06, 2018 1:03 am
by Harjas Sabharwal 1G
The answer to your question depends on the specific rate law for the reaction you are discussing. Generally if you plug in the change and then compare the resulting rate laws, you should be able to figure out what the change in rate is.

Re: Concentration and Rate Relationship

Posted: Tue Mar 06, 2018 10:57 am
by Christina Bedrosian 1B
for multiple reactions, see the effect of each reactant on the rate. so if one reactant is 1st order and another is 2nd order, the 1st would double the rate if the concentration is doubled and the 2nd one would quadruple the rate if the concentration was doubled. you usually view them separately like that